What is the integral from-1 to 1 of e^y-(y^2-5) ?

The integral from-1 to 1 of e^y-(y^2-5) is e-1/e+28/3

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integral from-1 to 1 of e^y-(y^2-5)

\int_{- 1}^{1} e^{y} - (y^{2} - 5) d y

e - \frac{1}{e} + \frac{28}{3}

Decimal Notation

11.68373 \dots

Solution steps

\int_{- 1}^{1} e^{y} - (y^{2} - 5) d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 1}^{1} e^{y} d y - \int_{- 1}^{1} y^{2} - 5 d y

\int_{- 1}^{1} e^{y} d y = e - \frac{1}{e}

\int_{- 1}^{1} y^{2} - 5 d y = - \frac{28}{3}

= e - \frac{1}{e} - (- \frac{28}{3})

Simplify

= e - \frac{1}{e} + \frac{28}{3}

\int_{1}^{5} (2 x - 1) d x

\int \frac{3 x}{2} d x

\int \frac{x ^{2}}{( 81 + x ^{2} ) ^{2}} d x

\frac{d}{d t} (ln (1 + t^{2}))

x \to 0 lim (ln (5) cos (x))

What is the integral from-1 to 1 of e^y-(y^2-5) ?
The integral from-1 to 1 of e^y-(y^2-5) is e-1/e+28/3